# Answer to Question #126533 in Mechanics | Relativity for Mwansa Kunda

Question #126533
Two objects are connected by a light string passing over a frictionless pulley. m1=5.00kg, m2=3.00kg, h=4.00m
The object of mass 5kg was released from rest. Using the principle of conservation of energy,
Determine;
i. The speed of the 3.0kg object just as the 5.0kg object hits the ground.
ii. The maximum height to which the 3.0kg object rises.
1
2020-07-20T15:13:03-0400 5 kg body locates 4 meters above ground. After it released it will go down because it heavier than 3 kg body. Energy conservation principle says:

"E_{p}=E_{k}"

(1) "m\\times a\\times h = \\frac{m\\times V^2}{2}"

where m - mass of body (5 kg), a - acceleration (not equals to gravitational acceleration in our case because we have 3kg body on other side), h - height (4 meters), V - velocity of body near ground.

acceleration could be found:

"F_{result} = F_{g5} - F_{g3}= m_{5}\\times g - m_{3} \\times g;"

"F_{result} = 5\\times 9.8 - 3\\times 9.8 = 49 - 29.4 = 19.6N";

"F_{result}=m_{5}\\times a;"

"a=\\frac{F_{result}}{m_{}}=\\frac{19.6}{5}=3.92 m\/s^2"

so, returning to formula (1) we can find V of body 5kg and body 3 kg (they a the same in this moment)

"V=\\sqrt{\\frac{m_{5}\\times a \\times h \\times 2}{m_{5}}}=\\sqrt{a \\times h \\times 2}=\\sqrt{3.92\\times 4\\times 2};"

"V=5.6m\/s"

Ansver for first question speed is 5.6 m/s

After 5 kg body hit ground, body 3kg will locate on height 8 m and continue move upward for some time, because it has some speed. To find maximum height which body 3 will have, let's use the same principle:

"m_{3}\\times g\\times h=\\frac{m_{3}\\times V^2}{2};"

"h=\\frac{V^2}{2\\times g}"

this time we use g because only force that affects body 3 kg is gravity.

"h=\\frac{5.6^2}{2\\times 9.8}=1.6m"

So maximal height of body 3 kg is 8 + 1.6 = 9.6m

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